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Repeating crossbow spring idea.

Post questions and info about pneumatic (compressed gas) powered cannons here. This includes discussion about valves, pipe types, compressors, alternate gas setups, and anything else relevant.
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Repeating crossbow spring idea.

Unread postAuthor: boyntonstu » Fri May 14, 2010 8:16 am

[youtube]http://www.youtube.com/watch?v=vqzoLga7lac&feature=related[/youtube]

Looks much easier to cock that my 50 pound pistol crossbow used to open a Ball Valve.

I would imagine that 80 to 100 pounds of force would open a BV pretty fast.

Calculating backwards:

A 3" radius ball valve handle needs to rotate in 0.004 seconds for the valve to open in .001 second. 90* is 1/4 rotation.

3.14 x 6" = 18.84"

18.84 / 4 = 4.71" (let's consider 5")

That works out to 5"/0.001 second or 5,000 inches per second.

That is 416 fps.

Half that is 208 fps or .002 seconds to open a BV 100%.

This is possible.
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Unread postAuthor: iknowmy3tables » Fri May 14, 2010 10:06 am

uhm I'm don't think that's how you calculate it, would the impedance of a ball valve really make the overall time only half as fast as the time for a light little
well you could solve using potential and kinetic energy I'll help you out with the calculation, it'll help practice for my physics exam. Just tell me the approximate mass, shape and dimensions of the parts in the rotating handle; your estimate of how much force static friction there is at the point where force is applied (do you understand what I mean? I just want to make sure that the force applies to the correct point of leverage); and maybe even maybe guesstimates on mass and length of a crossbow's bow (I might be able to check this on my crossbow at home but it's a 150lb bow)

also does anyone know if it's not okay to model the force on a bowstring using Hookes law?
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Unread postAuthor: jrrdw » Fri May 14, 2010 10:16 am

Didn't you build a dart launcher similar? Something was posted before along these lines.....

Cool toy non-the-less.
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Unread postAuthor: Hawkeye » Fri May 14, 2010 11:06 am

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Unread postAuthor: boyntonstu » Fri May 14, 2010 11:28 am

iknowmy3tables wrote:uhm I'm don't think that's how you calculate it, would the impedance of a ball valve really make the overall time only half as fast as the time for a light little
well you could solve using potential and kinetic energy I'll help you out with the calculation, it'll help practice for my physics exam. Just tell me the approximate mass, shape and dimensions of the parts in the rotating handle; your estimate of how much force static friction there is at the point where force is applied (do you understand what I mean? I just want to make sure that the force applies to the correct point of leverage); and maybe even maybe guesstimates on mass and length of a crossbow's bow (I might be able to check this on my crossbow at home but it's a 150lb bow)

also does anyone know if it's not okay to model the force on a bowstring using Hookes law?


That's how you calculate the speed required, not the forces or the resistances involved.

What I do know, is that a fish scale measures 2 pounds of force to move the BV handle.

Please calculate the force necessary to accelerate a 2 pound weight to 400 fps in .002 seconds.

Second calculation is a transfer of momentum necessary to get the 2 pounds to 200 fps.

Please try it.
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Unread postAuthor: Technician1002 » Fri May 14, 2010 12:08 pm

boyntonstu wrote:

That's how you calculate the speed required, not the forces or the resistances involved.

What I do know, is that a fish scale measures 2 pounds of force to move the BV handle.

Please calculate the force necessary to accelerate a 2 pound weight to 400 fps in .002 seconds.

Second calculation is a transfer of momentum necessary to get the 2 pounds to 200 fps.

Please try it.


Resistance does not equal mass. There will be a large conversion of momentum to heat in the resistance. A large portion of the applied force will be lost to drag. A friction measurement at rest might not be the same as the friction at higher speed.

The times for opening in the original post are for constant velocity of the rotation. The ~400 FPS is the average speed required, not the peak speed required. Due to the initial speed requirement of starting at a stop and ending at a stop, a very strong spring will be required to provide the high acceleration required to start, move, and safely stop the piston the 90 degrees in 1ms. I do not know of a spring capable of resonance at 500 Hz tied to a ball valve handle swinging 90 degrees. I do not know of a mechanical latch fast enough to release the spring at closed and catch it at open and back.

More important, I don't know of a ball valve that can sustain the forces necessary to do the job.
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Unread postAuthor: deathbyDWV » Fri May 14, 2010 4:34 pm

Not completely on topic but I recentlY got that same pistol crossbow with the self cocking lever for $25 from a website called "first rate security". The loading mech and stock he made are cool...
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Unread postAuthor: maverik94 » Fri May 14, 2010 6:04 pm

Wow. That repeating crossbow in the video is awesome!! I'm lovin his accent too :wink:
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Unread postAuthor: iknowmy3tables » Mon May 17, 2010 8:09 am

boyntonstu wrote:
iknowmy3tables wrote:uhm I'm don't think that's how you calculate it, would the impedance of a ball valve really make the overall time only half as fast as the time for a light little
well you could solve using potential and kinetic energy I'll help you out with the calculation, it'll help practice for my physics exam. Just tell me the approximate mass, shape and dimensions of the parts in the rotating handle; your estimate of how much force static friction there is at the point where force is applied (do you understand what I mean? I just want to make sure that the force applies to the correct point of leverage); and maybe even maybe guesstimates on mass and length of a crossbow's bow (I might be able to check this on my crossbow at home but it's a 150lb bow)

also does anyone know if it's not okay to model the force on a bowstring using Hookes law?


That's how you calculate the speed required, not the forces or the resistances involved.

What I do know, is that a fish scale measures 2 pounds of force to move the BV handle.

Please calculate the force necessary to accelerate a 2 pound weight to 400 fps in .002 seconds.

Second calculation is a transfer of momentum necessary to get the 2 pounds to 200 fps.

Please try it.

that's a way you can calculate things, if you take basic college level physics they will teach you how to calculate using either forces or by comparing the total energy of each system.
Tell me the mass, shape, and dimensions of the valve handle system because I need to calculate based on the total moment of inertia.
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